Hey DeeJay,
Thanks for detailed answer, it really helps as it shows the way I need
to follow!
It also helped me to realize that my question why
f1 f g x = mapMaybe f (g x)
has this type:
f1 :: (a -> b) -> (t -> Maybe a) -> t -> Maybe b
and not that (type which I expected):
f1 :: (a -> b) -> (t ->
Hi Dmitri,
your f1 function has 3 arguments, f, g and x.
you pass f as the first argument to mapMaybe, so it naturally must have type (a
-> b).
you pass the result of (g x) to the second argument of mapMaybe, so (g x) must
have type Maybe a. This means g must have the type (t -> Maybe a) where
I am trying to solve a problem from "The Craft of Functional Programming" book:
14.38 ... define the function:
data Maybe a = Nothing | Just a
composeMaybe :: (a -> Maybe b) -> (b -> Maybe c) -> (a -> Maybe c)
using functions:
squashMaybe :: Maybe (Maybe a) -> Maybe a
squashMaybe (Just (Just x)
